A metric space having a countable dense subset has a countable base.

Every separable metric space has countable base, where base is collection of sets {Vi} such that for any x that belongs to an open set G (as subset of X), there is a Vi such that x belongs to Vi.

2. Relevant equations

Hint from the book of Rudin: Center the point in a countable dense subset of the metric space and have a union of all "rational" radius.

3. The attempt at a solution

I need to understand this problem a bit more, so would appreciate any hints. I am mainly unclear as to why one needs to choose rational radius. Is it because rational numbere are countable? So we can have finitely many Vi.

My understanding so far is that if x belongs to G , then we need to prove that G is a union of Vi , where i belongs to N.

Now, based on the hint, we can find a point "p" in the dense subset A Intersection X, and have neighborhoods of all rational radii. Let any such neighborhood be Vi. Any neighborhood is an open set and the union of such open sets is open. Let this set be G , then any element that belongs to G , belongs to Union of Vi. But Vi is a collection of rational radii neighborhood and rational numbers are countable. Hence, the collection Vi is countable. In other words, {Vi} is the base.

The rational numbers are chosen precisely because they're countable. You can choose balls of radius 1/n for n in N if that makes you happy (and this is actually cleaner).

Then you said that we have finitely many V_i. Of course this is completely wrong. I'll let you figure out why.

Your understanding about G being a union of V_i's is correct. Although i doesn't have to belong to N: any union is OK (although by the nature of the collection of V_i's, any such union is going to be countable).

Your last paragraph isn't convincing. What you're supposed to do is start with a countable collection of open sets {V_i}. Then show that for any given open set G in X, we can choose a suitable subcollection of {V_i}, say {V*_i} such that G is the union of the V*_i's.